JEE Main Chemistry question
Cr(NH)Cl(aq) solution is titrated with HCl(aq): Cr(NH_3)_xCl_3(aq)+xHCl(aq)\rightarrow xNH_4^+(aq)+Cr^{3+}(aq)+(x+3)Cl^-(aq). 24.26 mL of 1.5 M HCl is used to titrate 1.580 g of Cr(NH)Cl. Thus x is
Answer: 6
Solution
Moles HCl mol moles complex. Molar mass of complex . So
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